Photoelectric effect?

Question

In a photoelectric experiment, it is found that a stopping potential of 1.00 V is needed to stop all the electrons when incident light of wavelength 260 nm is used and 2.30 V is needed for light of wavelength 207 nm. From these data, determine Planck's constant and the work function of the metal.

Answer 1

I think
E=hv

E=Vq

then
h= Vq/v since v=c/L
h= Vq/(c/L)
h=1.0 x 1.6202 E-19 / [2.99 E+8/260E-9]
h=1.393 e-34 J s ( gee this is close but no cigar. )

Actually h = 6.626 E-34 J s

Did I forgot something?

Actually in a photo effect we have a split of incident energy Ein into two namely Energy needed to remove an electron (hv’) [work function ]+ kinetic energy Ke of the emitted electron

Ein= hv’ + Ke
Ein= hv’ +(1/2)mV^2
V - in this case is velocity
v - is the threshold frequency for the incident photons,
v’ - is the threshold frequency for the photoelectric effect to occur,

Ein in both cases is the same
The Ke is determined by voltage bias Ke=Vq

Now we are ready do attempt the solution

hv1’ + Ke1= hv2’ + Ke2
h=(Ke2-Ke1)/( v1’ – v’2)

h= (V2q – V1q)/(c/L1 - c/L2)
h=(q/c)(V2 – V1)/(1/L1 - 1/L2)

Answer 2

Einstein's equation:
hf = W + eV
or hc/L = W + eV
now
hc/L1 = W + eV1 (1)
hc/L2 = W + eV2 (2)
(2) - (1)
hc(1/L2 - 1/L1) = e(V2 - V1)
so
h = e(v2 - V1)/[c(1/L2 - L1)]
= 1.6E-19 * 1.30/[3E8*(1/207E-9 - 1/260E-9)] = 7.04E-34 Js
the work function W = hc/L1 - eV1 = 6.52E-19 J